In triangle $ABC,$ $AP = BQ = \frac{PQ}{2}$ and $CR/RA = 3/2.$ Find the ratio of the area of triangle $BQC$ to the area of triangle $CRQ$

[asy]

pair A, X, Y, B, Z, C;

A = (0,0);

X = (0.3,-0.2);

Z = (0.4,0.2);

Y = 3*X;

B = 4*X;

C = 2.5*Z;

draw(X--Z--Y--C--B--A--C);

label("$A$",A,W);

label("$B$",B,S);

label("$C$",C,N);

label("$P$",X,SW);

label("$Q$",Y,SW);

label("$R$",Z,NW);

[/asy]

Please Try to Respond as Fast as possible.

Guest Feb 12, 2020

#1**+3 **

In triangle \(ABC\), \(AP = BQ = \frac{PQ}{2}\) and \(CR/RA = 3/2 \). Find the ratio of the area of triangle \(BQC\) to the area of triangle \(CRQ\).

Since AP = BQ = PQ/2. We can conclude that QB is 1/4 of AB.

Let the area of ABC = x

So BQC = \(\frac{x}{4}\)

Area of triangle CAQ = \(x-\frac{x}{4}=\frac{3x}{4}\)

Since CR/RA = 3/2, that means that CRQ = \(\frac{3x}{4}*\frac{3}{5}=\frac{9x}{20}\)

SO

Area of BQC is \(\frac{x}{4}\)

Area of CRQ is \(\frac{9x}{20}\)

Can you find the ratio now? (Hint: It's simple algebra!)

CalculatorUser Feb 12, 2020